How to Write NSE Technical Report

By

-

May 23, 2021

Table of Contents

The Nigerian Society of Engineers (NSE) was founded in the year 1958 to serve as an umbrella body for professional engineers in Nigeria. As a professional body, the NSE is dedicated to the professional development and enhancement of its members, and amongst other things, to upholding high ethics and standards in the practice of engineering profession in Nigeria.

Engineering graduates from COREN recognised universities who have practiced for a minimum of four years are qualified to apply to become corporate members of the NSE. During application, candidates are expected to provide documents such as 1st-degree certificate, NYSC certificate, statement of experience forms, acknowledgment form (from the branch), and must be endorsed by two financially up-to-date Registered Engineers who will act as the proposers. There are other different categories of membership in the NSE and the requirements for each cadre of membership are well specified on the NSE website.

Candidates following the B1/B2 route (described above) to corporate membership are required among other things to satisfy some minimum requirements which are assessed through technical reports, interviews, and examination. Candidates who successfully pass through this process can directly apply to COREN to be recognised as Registered Engineers without further exams or interviews.

One of the major hurdles of the NSE application process is to successfully write a detailed technical report containing your relevant work experience. The interesting aspect of this is that the report will be read and assessed by experienced registered engineers in the field of the candidate, who will eventually interview the candidate.

If the candidate is successful after this stage, he can proceed to write a computer-based test before the council takes a decision. Therefore, the aim of this article is to show you how to successfully write and defend your NSE technical report.

The NSE technical report is separated into Volume 1 and Volume 2. Volume 1 of the report contains the generalised post-graduate experience of the candidate, while Volume 2 is a technical report that contains specific engineering designs and/or construction that the candidate has executed under the supervision of a Registered Engineer.

In order to be successful, a candidate should be very sure that he is qualified to be admitted as a corporate member of the NSE. This is done by checking that he has met all the prescribed requirements, including a minimum of 4 years of professional work experience.

Engineering graduates who have met the four years postgraduate requirement but are not practicing engineering should not apply, since this will undermine the vision of upholding high professional standards in engineering in Nigeria. Only those who are practicing engineering should apply.

Therefore honesty, integrity, and confidence is very paramount in this case. Furthermore, it is very easy for experienced engineers to recognise someone who is not in active practise.

Guidelines for Writing Volume 1 – ‘Post-Graduate Experience’

The format or guideline for writing Volume 1 of the NSE technical report can be downloaded from the NSE website. This aspect of the technical report should contain the academic and work history of the candidate. The procedure for writing the academic history can be seen in the NSE guideline. However, I will elaborate on the work experience aspect.

The way the professional work experience of the candidate should be arranged as shown in the snippet below;

A candidate is expected to report on at least 5 projects he has been involved in since his graduation from the university. The projects can be in the form of design (consultancy) and/or construction. Candidates in academia can submit details of their research output (publications) that are relevant to the development of the engineering profession. The aim of the Volume 1 report is to show the depth of professional experience acquired by the candidate.

Post Graduate Training – Titled Experience

According to NSE guidelines, a candidate is expected to start by stating the title of the experience gained for each project. This will enable the examiners to know the direction that the candidate is going, and also know the minimum experience they are expecting the candidate to acquire from the project.

A civil engineering graduate who has been on a construction site as a site engineer should state ‘site supervision/management‘ as the experience gained. Someone who has been in the design office as a consultant can state the experience gained as ‘structural design‘. To be more specific, the candidate should add the actual nature of the design done since each project in the report is expected to be unique. Experience in the design of steel structures should be differentiated from experience in the design of reinforced concrete.

Therefore, the titled experience for each of the projects can be in the following forms (note: this is mixed up across different fields of engineering);

WORK EXPERIENCE 1: DESIGN OF SOLAR INVERTERS
WORK EXPERIENCE 2: DESIGN OF TIMBER BRIDGE
WORK EXPERIENCE 3: CONSTRUCTION SUPERVISION OF RIGID PAVEMENT
WORK EXPERIENCE 4: DESIGN OF EARTH DAMS
WORK EXPERIENCE 5: INSTRUMENTATION OF LPG GAS PLANT

For example, if you look at the titles above, the examiners will expect the candidate who stated the ‘Work Experience 2’ to have acquired a minimum basic experience in the design of timber bridges (not reinforced concrete or steel). With that, they are already looking out for some important details that will show that the experience gained is sufficient. All grey areas can be cleared during the interviews as the examiners will likely request to get more information from you.

Project Title

The title of the project where the experience was gained should be explicitly stated. The project title should be as written in the contract documents of the project. The following are examples of project titles;

(1) EXPANSION AND REHABILITATION OF ENUGU-ONITSHA EXPRESSWAY
(2) INSTALLATION OF 33 kVA TRANSFORMER FOR SILVA ESTATE, AKURE
(3) STRUCTURAL DESIGN OF RESIDENTIAL DEVELOPMENT FOR MR. AND MRS. GARBA BELLO
(4) DESIGN AND INSTALLATION OF HVAC FOR ORIENTAL HOTELS AND TOWERS

Organisation Name

The NSE technical report guidelines requires that you state the organisation/department where the experience was gained. For example;

FEDERAL MINISTRY OF WATER RESOURCES, NIGERIA (Dams and Reservoir Operations Department)
STRUCTVILLE INTEGRATED SERVICES, LIMITED
OVE ARUP AND PARTNERS (Project Management Department)
LAMBERT ELECTROMEC

Statement of Project Objective

No project starts in vain. As an engineer, it is expected that you should know the objective of the project that you are embarking on. The objective of a project should also determine whether you should agree to be part of the project or not. It is against NSE engineering ethics to be involved in projects whose aims are inimical to humanity and general development. For instance, projects linked to acts of terrorism should be rejected by professional engineers. Do not be like Tom Lehrer who made this quote;

Once the rockets go up, who cares where they come down? That is not my department.

The specific objectives of an engineering project could be;

(1) Construction of water treatment plant for raw water from Ogbese Reservoir for household distribution.
(2) Assessment of the sub-soil conditions of Lekki Peninsula with a view of obtaining the most appropriate foundation system for a medium-rise commercial development.

The objectives of the project from which you obtained your relevant experience can give the examiners an idea of the quality of experience and/or exposure that you must have gained.

Project Duration and Position Held

It is important to state the number of years or months you spent on each project. Remember as I said earlier, the aim of the examiners is to evaluate the depth of experience you have acquired in the profession. Nobody expects you to know it all, but in the areas where you have worked, you are expected to be good at it in order to be called an engineer.

Furthermore, the job title or position you held during your engagement should be stated. Job titles are usually provided in your employment letter and should be very professional titles that reflect the nature of your responsibility. When the actual job title is generic, you can add the specific role or responsibility in parenthesis. Some examples of valid job titles are;

(1) GRADUATE RESEARCH ASSISTANT (Department of Mechanical Engineering)
(2) PUPIL ENGINEER (Structural Design)
(3) ASSISTANT SITE SUPERVISOR
(4) GRADUATE INTERN (Facility Management), etc

It is awkward for you to handle a project from start to finish immediately after graduation without the supervision of a Registered Engineer. Therefore, you should be properly guided when writing about your experience.

Description of Experience

This is an important aspect of the NSE ‘Volume 1’ technical report. In this section, you are expected to describe the project, your level of participation in the project, your responsibility, the summary of your daily routine/job description, challenges encountered during the project, solutions proferred, and lessons learned from the project.

It is very important for you to write on projects that you truly worked on. Avoid giving false reports or giving your technical report to someone else to write for you. You should own your report completely. By so doing, you will be able to defend all the processes and operations of the project from start to finish. This should include the design, construction methodology, and construction management. Lies and discrepancies can be easily detected since you are going to be examined by engineers who do exactly the same thing.

Furthermore, even if you worked on construction aspects only, you should have an idea about how the design is done. This is what makes you a complete engineer.

In order to make this section clear, you should break it down into sub-sections as recommended in the NSE technical report guideline. The ‘description of experience report’ should be written in first person singular or plural.

If you have truly participated in an engineering project, you should have no challenge in telling the story of what you did during the project. The only difference is that you will have to tell the story in a technical way. Having said that, you should use technical terms in your description of daily work activities.

For civil engineers on site, it is very common to use expressions that are easily understood by the artisans. For instance, terms like ‘shoot-out‘ (for cantilevers), ‘iron 16‘ (for Y16 mm rebars), ‘pillars‘ (for columns), etc should be avoided in your report. Also, the proper description of each item of work such as ‘preparation and installation of formwork‘, ‘pouring/casting of concrete‘, ‘creation of diversion‘, ‘establishment of construction levels‘, etc should be used.

Every project is unique and has its own challenges. A good engineer will always reflect on the challenges encountered during a previous project and how it was resolved. This is because experience is what makes a complete engineer. You can as well think of how to improve on the solutions you adopted in the past. Some examples of challenges that can be encountered on site are;

(1) Groundwater control problems
(2) Human resources management problems
(3) Discrepancies between design specifications and actual site conditions
(4) Construction cost management, etc

In order to assess your level of exposure in the engineering profession, the examiners will like to know the challenges you encountered and how you solved them. Remember that your storytelling should be technical, and the solutions you proffered should make engineering sense (technical and management wise).

Even if you had made wrong decisions, there is no problem since you are expected to state the lessons you learned from each project. We all make mistakes, but the lessons you learn from them are more important.

Pictures

Ultimately, you are expected to back up each project with pictures to add more weight to your claims. So always take pictures of your construction projects. It is very important.

Guidelines for Writing Volume 2 – ‘NSE Technical Report on Two Selected Projects’

In Volume 2 of the NSE technical report, you are expected to select two of your best projects from the five listed in Volume 1 and write a detailed report on them. Preferably, the two projects selected should not be related and should offer a completely different experience to the candidate.

For instance, if the first project report is on the design of a reinforced concrete residential building, the second should be preferably come from say, design and construction of an industrial steel building, shoreline protection, retaining walls and culverts, bridges, construction of jetties/wharves, tank farms, and reservoirs, etc.

Candidates with experience in different fields can report on highway construction, water treatment and supply, etc.

According to NSE technical report guidelines, Volume 2 should be divided into three chapters. The first and second chapters should contain reports on the first and second projects respectively, while the third chapter should contain the recommendation and conclusion.

The arrangement of chapters 1 and 2 should be as follows;

1.0 INTRODUCTION
1.1 JUSTIFICATION OR NEED FOR PROJECT
1.2 PRELIMINARY STUDIES/INVESTIGATIONS
1.3 DESIGN CONSIDERATIONS/CRITERIA
1.4 STANDARDS AND SPECIFICATIONS
1.5 METHODOLOGY AND DESIGN CALCULATIONS
1.6 DRAWINGS
1.7 PREPARATION OF BEME (INCLUDING TAKE–OFF SHEETS)
1.8 CONSTRUCTION/INSTALLATION/ANALYSIS/TEST & CALCULATIONS
1.9 ANALYSIS OF TEST RESULTS AND COMMISSIONING
1.10 PROJECT OUTCOME

Introduction

In this section, the project should be properly introduced. The introduction should include the title of the project, the client, the consultants, and the contractors. Furthermore, the location of the project should also be included. For buildings, the features of the building should be included such as the materials to be used for the construction (reinforced concrete, structural steel, timber, etc), the number of floors, the floor area, etc. Other specific features such as suspended swimming pools, helipad, tanks, etc should also be included.

Every feature that can give an idea of the magnitude of the selected project and the expected level of construction difficulty should be stated.

Justification of the Project

The exact need of the project should be stated. This could be stated in form of;

Construction of a two-lane flyover bridge at Eleme Junction to ease traffic congestion
Construction of public swimming pool for residents of Springhill Estate for recreation and relaxation purposes
Construction of Okoja motor park to provide parking facility for commuters, and to avoid on-street parking

Preliminary Studies/Investigations

All the preliminary studies carried out before the commencement of the project should be stated. These can include the Environmental Impact Assessment (EIA), site surveys, geotechnical site investigation, wind funnel tests, etc. The findings of these investigations and how they affect the project in terms of planning, design, and execution should also be stated.

Design Considerations/Criteria

The candidate is expected to state the design considerations for the project. These can include the decisions taken from the preliminary studies or site investigation. The exposure conditions of the proposed structure, the basic wind speed, potentials for differential settlement or temperature difference should be stated.

Furthermore, the anticipated direct actions on the structure should be stated such as ‘highway bridge to be subjected to abnormal traffic‘, ‘building to be used as a place of worship‘, ‘building founded on a water logged area‘ etc. All considerations that will affect the design of the building should be stated.

Standards and Specifications

The code of practice or standard adopted in the design should be clearly stated.

Methodology/Design Calculations

For reports on construction, the construction methodology should be stated. For design works, the design calculation should be shown in full. Remember that the design calculation sheet for NSE technical report should be presented in the standard format of ‘Reference – Calculation – Output‘.

Drawings

All design drawings (structural detailing) should be included. Drawings should be presented using the generally accepted format and scales. A lot of people have been disqualified from NSE exams for presenting shoddy drawings. All drawings should be clean, legible, and properly formatted. Remember to show plan and sections (for slabs), and elevation and sections (for beams and columns).

Preparation of BEME

The Bill of Engineering Measurement and Evaluation (BEME) for each project should be included. This should also include the quantity take-off sheet. As far as what we know is concerned, BEME is concerned with the calculation of quantities and cost of items of work in an engineering project that requires engineering judgement in the process. Otherwise, it something that should be left to quantity surveyors in my opinion.

Therefore for civil engineering projects, I will recommend limiting your BEME to just concrete and reinforcements works (or structural steel and the accessories). Calculation of quantities such as plaster, blocks/bricks, doors, windows, tiles, etc is not necessary in my opinion.

Earth Dams: Types, Construction, and Modes of Failure

By

Ubani Obinna

-

May 20, 2021

Table of Contents

Dams are structures that are constructed to impound a water body such as a stream or a river. The upstream (reservoir) of a dam is crucial for water storage which can be used for irrigation, municipal water supply, hydropower generation, flood control, fishing, and recreation. There are different types of dams such as earth dams, gravity dams, arch dams, etc.

Earth dams are dams that are constructed using natural materials such as natural soils, rocks, clays, and gravel. It is the most ancient type of embankment and can be constructed using familiar processes and primitive equipment. Unlike gravity and arch dams which require a sound foundation and more complex construction materials and methods, earth dams can be founded on natural soils. However, they are more susceptible to failure when compared with other types of dams.

Types of Earth Dams

There are three popular types of earth dams and they are;

Homogeneous Embankment type
Zoned Embankment type
Diaphragm type

Homogeneous Embankment Type

This is the simplest type of earth dam. It is constructed using a single material (same type of soil) and hence can be considered to be homogeneous throughout. To aid water tightness and stability, a blanket of relatively impervious material may be placed on the upstream face. This type of embankment is attractive when only one type of material is economically or locally available. However, this type of earth dam is more suitable for low to moderately high dams and for levees. Large dams are seldom designed as homogeneous embankments.

Seepage can be a major problem of purely homogenous earth dams. As a result, huge sections are usually required to make it safe against piping, stability, etc. To overcome this problem, it is usually very typical to add an internal drainage system such as a horizontal drainage filter, rock toe, etc. The internal drainage system keeps the phreatic line (i.e. top seepage line) well within the body of the dam, and steeper slopes and thus, smaller sections can be used, the internal drainage is, therefore, always provided in almost all types of embankments.

homogenous earth dam with drainage filter — **Fig 2***: Homogeneous type embankment earth dam* with drainage filter

Zone Embankment Type

Zoned embankments are usually provided with a central impervious core, covered by a comparatively pervious transition zone, which is finally surrounded by a much more pervious outer zone. The central core checks the seepage. The transition zone prevents piping through cracks which may develop in the core. The outer zone gives stability to the central impervious fill and also distributes the load over a large area of foundations.

zoned embankment type of earth dam — **Fig 3***: Zoned embankment earth dam*

This type of embankment is widely constructed and the materials of the zones are selected depending upon their availabilities. Compacted clay can be used for the central impervious core. The clay material should be carefully selected and should have a coefficient of permeability of less than 1 x 10^-9 m/s irrespective of the compaction energy applied. Furthermore, in order to avoid desiccation-induced shrinkage cracks, the volumetric shrinkage should not exceed 4% and the unconfined compression strength (UCS) should exceed 200 kN/m².

Freely draining materials, such as coarse sands and gravels, are used in the outer shell. Transition filters are provided between the inner zone. This type of transition filter is always provided, whenever there is an abrupt change of permeability from one zone to the other.

Diaphragm Type Embankment

Diaphragm type embankment earth dam has a thin impervious core, which is surrounded by earth or rock fill. The impervious core, called the diaphragm, is made of impervious soils, steel, timber, concrete, or any other materials. It acts as a water barrier to prevent seepage through the dam.

The diaphragm may be placed either at the center as a central vertical core or at the upstream face as a blanket. The diaphragm must also be tied to the bedrock or to a very impervious foundation material. This is to avoid excessive under-seepage through the foundation.

DIAPHRAGM TYPE OF EARTH DAM — **Fig 4***: Diaphragm type embankment*

The diaphragm type of embankment is differentiated from zoned embankments, depending upon the thickness of the core. If the thickness of the diaphragm at any elevation is less than 10 meters or less than the height of the embankment above the corresponding elevation, the dam embankment is considered to be of “Diaphragm Type”. But if the thickness equal or exceeds these limits, it is considered to be zoned embankment type.

Methods of Construction

There are two methods of constructing earthen dams:

Hydraulic-fill Method; and
Rolled-fill Method.

Hydraulic-fill Method

In this method of construction, the dam body is constructed by excavating and transporting soils by using water. Pipers called flumes, are laid along the outer edge of the embankment. The soil materials are mixed with water and pumped into these flumes. The slush discharged through the outlets in the flumes at suitable intervals along their lengths. The slush, flowing towards the centre of the bank, tends to settle down. The coarser particles get deposited soon after the discharge near the outer edge, while the fines get carried and settle at the centre, forming a zoned embankment having a relatively impervious central core.

Since the fill is saturated when placed, high pore pressures develop in the core materials, and the stability of the dam must be checked for these pressures. This type of embankment is susceptible to settlement over a long period, because of slow drainage from the core. Hydraulic-fill method is, therefore, seldom adopted these days, Rolled-fill method for constructing earthen dams is, therefore, generally and universally adopted in these modern days.

Rolled-fill Method

The embankment is constructed by placing suitable soil materials in thin layers (15 to 30 cm) and compacting them with rollers. The soil is brought to the site from burrow pits and spread by bulldozers, etc. in layers. These layers are thoroughly compacted by rollers of designed weights. Ordinary road rollers can be used for low embankments (such as for levees or bunds); while power-operated rollers are to be used for dams. The moisture content of the soil fill must be properly controlled. The best compaction can be obtained at a moisture content somewhere near the optimum moisture content.

Failure of Earth Dams

Earth dams are less rigid and hence more susceptible to failure. Every past failure of such a dam has contributed to an increase in the knowledge of the earth dam designers. Earthen dams may fail, like other engineering structures, due to improper designs, faulty constructions, lack of maintenance, etc. the various causes leading to the failure of earth dams can be grouped into the following three classes.

Hydraulic failures
Seepage failures
Structural failures.

These causes are describes below in details:

Hydraulic Failures

About 40% of earth dams failures have been attributed to these causes. Hydraulic failure of earth dams can occur due to over-topping of the top of the dam, erosion of the upstream face, erosion of the downstream face due to the formation of gullies, and erosion of the downstream toe.

Seepage Failures

Controlled seepage or limited uniform seepage is normal in all earth dams, and ordinarily it does not produce any harm. However, uncontrolled or concentrated seepage through the dam body or through its foundation may lead to piping or sloughing and the subsequent failure of the dam. Piping is the progressive erosion and subsequent removal of the soil grains from within the body of the dam or the foundation of the dam. Sloughing is the progressive removal of soil from the wet downstream face. More than 1/3rd of the earth dams have failed because of these reasons.

Structural Failures

About 25% of the dam failures have been attributed to structural failures. Structural failures are generally caused by shear failures, causing slides. This is majorly an issue of slope stability and foundation stability of the dam.

Detailing of Reinforced Concrete Slabs

By

Ubani Obinna

-

May 14, 2021

Table of Contents

Structural detailing is the process of interpreting design information and instructions using drawings and schedules. In reinforced concrete slabs and other structures, detailing entails using drawings and schedules to specify the dimensions and arrangement of structural members, material properties, clear cover, reinforcement sizes, spacings, and arrangement.

It is the duty of the Designer and the Detailer to ensure that the information provided in the drawing is correct since the same will be used for execution on site. The essence of this article is to provide information on the detailing standards for reinforced concrete slabs according to the requirements of the Eurocodes and UK practice.

Detailing Information

The design information that should be provided in the detailing of reinforced concrete slabs include:

Layout and section drawings including details of holes and upstands, etc.
Concrete grade and aggregate size (minimum standard 25/30 MPa and 20mm).
Nominal cover to reinforcement and controlling design consideration, fire or durability (standard 20mm for internal conditions 40mm for external conditions).
Main reinforcement bar runs and positions. This should include:
- diameter, pitch of bars, and location (e.g. T1, T2, B1, B2, etc.)
- type of reinforcement and bond characteristics (standard: H)
- fixing dimensions to position bar runs and ends of bars.
Details of any special moment bars connecting slab to wall or column.
Details of cut-off rules, if other than standard shown in Model Details.
Details of fabric required. For coffered slabs, this should include the fabric required in the topping and in the bottom of solid sections around columns. Sufficient details should be given to show that the reinforcement will fit in the depth available allowing for laps in the fabric. Guidance should be given for the additional area required for laps otherwise 22% will be assumed for 300mm laps.
Details of insertions, e.g. conduit, cable ducting, cladding fixings, etc., should be given where placing of reinforcements is affected.

The Minimum Area of Reinforcement for Solid Slabs

According to Clauses 9.3.1.1, 9.3.1.2 and 9.2.1.1 of EC2;

Tension reinforcement:
- A_s,min = 0.26b_tdf_ctm/f_yk ≥ 0.0013b_td where:
  - b_t is the mean width of the tension zone
  - d is the effective depth
  - f_ctm is determined from Table 3.2 of EC2
  - f_yk is the characteristic yield strength
This also applies to nominal reinforcement.
Minimum bottom reinforcement in direction of span: 40% of the maximum required reinforcement.
Minimum top reinforcement at support (e.g. where partial fixity exists): 25% of the maximum required reinforcement in span, but not less than A_s,min. This may be reduced to 15% for an end support.
Secondary transverse reinforcement: 20% of main reinforcement except where there is no transverse bending (e.g. near continuous wall supports).
Preferred minimum diameter of reinforcement for solid slabs: 10mm.

The area of bottom reinforcement provided at supports with little or no end fixity assumed in design should be at least 0.25 that provided in the span.

Bar spacing

According to Clauses 8.2 and 9.3.1.1 of EC2, the recommended minimum spacing of reinforcing bars is 75 mm and 100 mm for laps.

Maximum spacing of bars for slabs
• Main bars: 3h ≤ 400mm (in areas of concentrated loads 2h ≤ 250mm)
• Secondary bars: 3.5h ≤ 450mm (in areas of concentrated loads 3h ≤ 400mm)

Where h is the thickness of the slab.

Anchorage and Lapping of Bars

For high yield and 500 Grade steel, the table below gives typical anchorage and lap lengths for ‘good’ and ‘poor’ bond conditions. For ends that are on ‘direct supports’ the anchorage length beyond the face of the support may be reduced to d but not less than the greater of 0.3 l_b,rqd, 10b or 100mm.

Table 1: Typical anchorage and lap length for solid slabs

Where loading is abnormally high or where point loads are close to the support, reference should be made to EC2, Sections 8 and 9. Lap lengths provided (for nominal bars, etc.) should not be less than 15 times the bar size or 200 mm, whichever is greater. The arrangement of lapped bars should comply with Figure 1;

lapping of bars in supports — **Figure 1:** Guideline for lapping in solid slabs

Simplified Curtailment Rules for Reinforcement

When only the minimum percentage of reinforcement is provided, there should be no curtailment when detailing reinforced concrete slabs. Simplified rules for curtailment of bars may be used without bending moment diagrams, provided adjacent spans are approximately equal (within 15%) and provided that the loading is uniformly distributed. The simplified rules for curtailment in solid slabs can be seen in Figures 2 to 5.

At internal supports in one-way and two-way slabs, the top reinforcement should extend into the span by 0.3 x times the length of the span as shown in Figure 2.

slab curtailment rules — **Figure 2**: Curtailment rules for top reinforcements in solid slabs

When the end support of a solid one-way or two-way slab is completely restrained (for example, when a solid slab is supported by a shear wall), the bars should be returned into the span by 0.3 x span as shown in Figure 3.

curtailment rule restrained end support 1 — **Figure 3:** *Curtailment rules for restrained end support*

For external unrestrained support (for example, slabs supported by masonry walls), the bottom reinforcement should be returned by 0.1 x span. Where partial fixity exists (for example, end support of a slab supported by beams), the bottom reinforcement should be returned by 0.15 x span.

**Figure 3:** Curtailment rules for unrestrained end support of slabs

At cantilevers, the main top reinforcements should extend into the span by at least 1.5 x times the length of the cantilever or 0.3 times the length of the span whichever is greater. It is also recommended to provide at least 50% of the top reinforcement at the bottom in order to help control deflection.

detailing of cantilever slab — **Figure 4:** Curtailment rules for cantilevers

In other circumstances, the curtailment of the main longitudinal reinforcement should be related to the bending moment/shear force diagrams.

Notation for Locating Layers of Reinforcement

Reinforcement is fixed in layers starting from the bottom of the slab upwards and bar marks should preferably follow a similar sequence of numbering.

Notation is as follows:

• abbreviation for top outer layer T1 (or TT)
• abbreviation for top second layer T2 (or NT)
• abbreviation for bottom second layer B2 (or NB)
• abbreviation for bottom outer layer B1 (or BB)

bar layer notation in slabs — **Figure 5**: Notation for reinforcement layers

Reinforcement Bars and Indicator Lines

In slab detailing, every reinforcement bar is assigned a bar mark. Each bar mark is unique to a type of reinforcement, grade, size, dimensions, and shape. Therefore a bar mark can represent a single bar or a group of bars. Every bar mark is represented on plan by a typical bar drawn to scale, using a thick line (generally, rebar lines should be thicker than all other lines in the detailing drawing).

The reinforcement bar is positioned approximately midway along its indicator line (also called the call-out line), the junction of the bar and the indicator line is highlighted by a large dot. The first and last bars in a zone of several bars are represented by short thick lines, their extent indicated by arrowheads. Bends or hooks, when they occur at either end of the typical bar are represented by a medium dot or similar as shown in Figure 7(b).

Sometimes, hooks or bent bars are drawn on plan as though laid flat (see Figure 7a). This is actually the commonest method of detailing. However, confusion on site can result if some of these bars are required to be fixed flat and some upright. Sections and notes should be provided to clarify this method if used.

Example

typical simply supported slab panel — **Figure 6**: Typical simply supported slab panel

Let us consider the slab panel shown above with simply supported assumptions. The reinforcement specified in all directions is H12 @ 200mm spacing.

To calculate the number of rebars to be provided in each direction, the following steps can be followed;

In the short span direction;
n = [(l_x – b_w – p)/p] + 1 = [(3105 – 230 – 200)/200] + 1 = 14.375 (provide 15 Nos of H12 bars)
Note: p is the spacing of the bars and it is assumed that the laying of slab mat reinforcement starts at 0.5p from the face of the beam.

Similarly in the long span direction;
n = [(l_y – b_w – p)/p] + 1 = [(3470 – 230 – 200)/200] + 1 = 16.2 (provide 17 Nos of H12 bars)
Note: On site, you will eventually have a spacing that is slightly less than 200 mm center to centre (about 178 mm, which is good/conservative). But if you provide 16 numbers, you will have spacing greater than 200 c/c which is not too good.

Detailing of reinforced concrete slab — **Figure 7**: **(a)** Hooks/bends drawn on plan as though laid flat **(b)** Hooks/bends represented by a medium dot

In figure 7(a) the bend/hook is drawn in plan as flat, while in figure 7(b) it is represented using a medium dot. Either method is acceptable in standard, but in Nigeria, the former is more popular.

When there are multiple zones/panels with similar bar marks, the number of bars in each panel can be written and the total summed up in the call out. This saves time and paper space in the detailing of reinforced concrete slabs. An example is shown in Figure 8;

SLAB DETAIL 3 — **Figure 8**: Detailing similar bar marks in multiple zones/panels

When there are serious space restrictions on the paper, the calling up of bars can be written along the indicator lines as shown in Figure 9. In extreme cases, it can be written along the bars.

alternative forms of detailing — **Figure 9**: Reinforcement call up written along the indicator lines in the slab

When bars are to be detailed in a panel/zone that is varying in dimension, the approach shown in Figure 10 should be used;

varying bar mark — **Figure 10**: Detailing of reinforcement in a zone with varying dimension

Edge reinforcement

According to Clauses 9.3.1.4 of EC2, reinforcement should be placed along free (unsupported) edges of slabs and at corners that are supported on both sides. This allows the distribution of local loads which helps to prevent unacceptable cracking. This reinforcement may be supplied in the form of U-bars as shown in Figure 11.

Numerical Integration for Engineers

By

Ubani Obinna

-

May 9, 2021

Table of Contents

Generally, integration is the process of summing up slices or parts in order to find the whole. If dx represents a small displacement or change along the direction x, the process of integration will give a function g(x), the derivative of which △g(x) is equal to the function f(x). This is indicated by the integral sign ∫. Thus, ∫f(x)dx is the summation of the product of f(x) and dx. Numerical integration is a computational (approximate) approach of evaluating definite integrals.

A definite integral is defined by limits (say a and b) and it is given by;

\int_{a}^{b} f(x) \,dx

Numerical integration has a lot of applications in engineering such as in the computation of areas, volumes, and surfaces. It also has the advantage of being easily programmable in computer software. In civil engineering, it can be applied in the computation of earthwork volumes such as cut and fill in road construction.

Read Also…
Linear Interpolation for Engineers

For the approximation of definite integrals of the form ∫f(x)dx, the numerical quadrature is normally used. In this method, the function f(x) is normally replaced with an interpolating polynomial p(x) which on integration, obtains an approximate value of the definite integral.

To obtain the numerical solution of functions using the numerical quadrature, the first step is usually to select a distinct set of equally spaced nodes {x₀, x₁, …., x_n} from the interval [a,b]. The smaller the spacing of the nodes or intervals, the more accurate the solution.

Let us assume that P_n is the lagrange interpolating polynomial then we can write;

\\P_n(x) = \sum_{i=0}^{n} f(x_i)L_i(x)

If we integrate P_n excluding its truncation error term over [a ,b], we obtain;

\int_{a}^{b} f(x) \,dx\ =\int_{a}^{b} \sum_{i=0}^{n} f(x_i)L_i(x),dx\\= \sum_{i=0}^{n} a_if(x)

Where a_i= ∫L_i(x)dx for i = 0, 1, 2, 3, … n

There are many methods of approximating the numerical quadrature for numerical integration but we are going to consider the most popular ones which are;

Trapezoidal Rule, and
Simpson’s rule

The Trapezoidal Rule for Numerical Integration

The Trapezoidal rule for numerical integration is obtained from considering the integration formula produced by using first Lagrange polynomials with equally spaced intervals. To evaluate ∫f(x)dx within the limits [a, b], let x₀ = a and x₁ = b. Then h = b – a = x₁ – x₀

Using the linear Lagrange polynomial;

P₁(x)= [(x – x₁)/(x₀ – x₁)]f(x₀) + [(x – x₀)/(x₁ – x₀)]f(x₁)

Therefore;

\int_{a}^{b} f(x) \,dx\ =\int_{a}^{b}[((x - x_1)/(x_0-x_1))f(x_0) + ((x - x_0)/(x_1-x_0))f(x_1)]dx

The equation above eventually yields;

∫f(x)dx = (x₁ – x₀)/2[f(x₁) + f(x₀)}

But (x₁ – x₀)/2 = h
Hence,

\int_{a}^{b} f(x) \,dx\ = h/2[f(x_1) + f(x_0)] ---- (2.3)

Equation (2.3) is known as the Trapezoidal rule.

It gives good approximation to the value of ∫f(x)dx when the curve of y = f(x) when the interval [a, b] is small, and deviates slightly from the trapezium aABb. However, in a situation when the deviation is violent, i.e. the interval [a, b] is very large, the accuracy in the approximation to the value of ∫f(x)dx can be improved by dividing the interval [a, b] into a larger (even) number of trapezoid of smaller width. This is referred to as the Composite Trapezoidal Rule.

Composite Trapezoidal Rule

Under the condition that warrants the use of composite trapezoidal rule, we can establish the general formula using the figure below;

Trapezoidal rule of numerical integration

The area;

A₁ = h/2(y₀ + y₁)
A₂ = h/2(y₁ + y₂)
A₃= h/2(y₂ + y₄)
.
.
.
A_n = h/2 (y_n-1 +y_n)

Hence
∑A_i = (A₁ + A₂ + A₃ + ⋯ + A_n)
= h/2 [y₀ + y_n+ 2(y₁ + y₂ + y₃ + ⋯ + y_n-1]

The generalised Trapezoidal Rule can therefore be expressed as;

\int_{a}^{b} f(x) \,dx\ = h/2[f(x_0) + f(x_n) + 2\sum_{i=1}^{n-1}f(x_i)] ---- (2.4)

The Simpson’s Rule for Numerical Integration

If we consider the integration formula derived by using the second Lagrange polynomials with equally spaced intervals. If the function f(x) is replaced by an arc of a parabola, and the origin temporarily shifted to a point x = x₀ by putting x = X + x₀ . We may therefore write the equation of the parabola A₀A₁A₂ as y = y₀ + bX + cX².

The area of the two adjacent strips under A₀A₁A₂ is approximately;

\int_{0}^{2h} (y_o + bX + cX^2)dx\ = 2h(y_o + bh + 4/3ch^2)

Ultimately;

\int_{0}^{2h} (y_o + bX + cX^2)dx\ = h/3(y_o + 4y_1 + y_2)

Composite Simpson’s Rule

Similar to Trapezoidal rule, the accuracy of Simpson’s rule can be improved by dividing the interval [a, b] into a larger (even) number of strips of smaller width. Let us evoke fig 6.2, and by considering two successive strips at a time, we can write the expression of Simpson’s formula as thus:

∫f(x)dx = h/3[y₀+ 4y₁ + y₂) + (y₂ + 4y₃ + y₄) + … + (y_n-2 + 4y_n-1 + y_n)

In a more compact form;

∫f(x)dx = h/3[y₀ + y_n + 2∑even ordinates + 4∑odd ordinates]

Worked Example

Evaluate the integral below using 10 intervals;

\int_{0}^{1} (sinx)dx\

Solution
At 10 intervals, h = 0.1

y₀ = f(x₀) = sin(0) = 0
y₁ = f(x₁) = sin(0.1) = 0.0998 (radians)
y₂= f(x₂) = sin(0.2) = 0.1986
y₃ = f(x₃) = sin(0.3) = 0.2955
y₄ = f(x₄) = sin(0.4) = 0.3894
y₅ = f(x₅) = sin(0.5) = 0.4794
y₆ = f(x₆) = sin(0.6) = 0.5646
y₇ = f(x₇) = sin(0.7) = 0.6442
y₈ = f(x₈) = sin(0.8) = 0.7173
y₉ = f(x₉) = sin(0.9) = 0.7833
y_n = f(x₁₀) = sin(1.0) = 0.8414

Using the composite Trapezoidal rule;
∫f(x)dx = h/2[y₀ + y_n+ 2(y₁ + y₂ + y₃ + ⋯ + y_n-1]
∫sin(x)dx = 0.1/2[0 + 0.8414 + 2(0.0998 + 0.1986 + 0.2955 + 0.3894 + 0.4794 + 0.5646 + 0.6442 + 0.7173 + 0.7833] = 0.4593

Using the composite Simpson’s rule;
∫f(x)dx = h/3[y₀ + y_n + 2∑even ordinates + 4∑odd ordinates]
∫sin(x)dx = 0.1/3[0 + 0.84814 + 2∑(0.1986 + 0.3894 + 0.5646 + 0.7173) + 4∑(0.0998 + 0.2955 + 0.4794 + 0.6442 + 0.7833)] = 0.4596

The exact solution

\int_{0}^{1} (sinx)dx\ = 0.459697

Therefore, it can be seen that the composite Simpson’s rule approximates better than the composite Trapezoidal rule.

Free-Standing Sawtooth Staircase | Analysis, Design, and Detailing

By

Ubani Obinna

-

May 8, 2021

A free-standing sawtooth staircase is a type of slabless (without waist) staircase that is freely supported at the landing. By implication, this staircase comprises of the thread and risers only, which are usually produced using reinforced concrete.

Analytical and detailing solutions exist for reinforced concrete sawtooth and free-standing staircases, but when the two systems are combined, there may be a challenge with the detailing due to the well-known structural behaviour of cantilevered type structures.

**Fig 1:** Typical section of a free-standing sawtooth staircase

In cantilevers, the main reinforcements are provided at the top (the tension area), and furthermore, they must be properly anchored and/or extended into the back-span for good anchorage/development length, and to resist the hogging moment that exists at the back of the cantilever.

In reinforced concrete slabs, the cantilever reinforcement should extend into the back-span by at least 0.3 times the span of the back-span, or 1.5 times the length of the cantilever, whichever is greater. The same requirement of reinforcements extending cantilever reinforcements applies to reinforced concrete beams too.

It can therefore be seen that continuity of reinforcements is an important detailing requirement of cantilevered type structures. Without the introduction of haunches, this may be difficult to achieve in free-standing sawtooth staircases.

In sawtooth staircases, the main reinforcements are provided on the landing, which is connected to the risers using links. A typical detailing sketch of a sawtooth staircase is shown below;

typical detailing of sawtooth staircase — **Fig 2:** Typical detailing sketch of sawtooth (slabless) staircase

Knowing full well that this type of staircase has been successfully designed and constructed (see Fig. 3). The pertinent questions to ask are, therefore;

(1) What is the detailing procedure of a free-standing sawtooth staircase without the use of haunches?
(2) Is the continuity of reinforcements necessary at the landing-riser junction?
(3) Are the links rigid enough to provide the needed continuity?

engenharianerd CObh63DjPL — **Fig 3:** Well constructed free-standing sawtooth staircase

To provide an insight to the answers, let us consider a finite element model of a free-standing sawtooth staircase.

The properties of the staircase are as follows;

Width = 1000 mm
Width of landing = 1000 mm
Height of riser = 175 mm
Width of thread = 275 mm
Thickness of all elements = 150 mm

Loading
(1) Self weight
(2) Finishes of 1.2 kN/m²
(3) Imposed load of 3 kN/m²

Load Combination
ULS = 1.35g_k + 1.5q_k

finite element model of free standing sawtooth staircase — **Fig 4:** Finite element model of the free-standing sawtooth staircase

Some of the analysis results are given below;

Fig 5: Deflection profile of the staircase under uniformly distributed load

**Fig 6:** Transverse bending moment of the staircase (ULS)

**Fig 7:** *Longitudinal bending moment of the staircase (ULS)*

Torsion — **Fig 8:** *Twisting moment of the staircase (ULS)*

From the nature of the internal stresses distribution, is it safe to say that the true cantilever behaviour of structures is not properly represented in free-standing sawtooth staircases? From the deflection profile, this is probably not the case. Can we get sketches of typical detailing guide from you?

Thank you for visiting Structville today. God bless you.

Design of Precast Columns | Worked Example

By

Ubani Obinna

-

May 4, 2021

Table of Contents

Precast concrete columns are reinforced concrete columns that are cast and cured on the ground before being hoisted up and installed in their desired positions. Just like in-situ columns, precast columns are capable of resisting shear, axial force, and bending moment, however, careful attention must be paid to their connection details. The design of precast concrete columns involves the provision of adequate member size, reinforcement, and connection details to satisfy internal stresses due to externally applied loads, second-order effects, and lifting.

Different connection conditions can be adopted by different manufacturers. The foundation connection of a precast column may be achieved by allowing reinforcement bars to project from the column which is then passed through established sleeves before being filled with concrete grout. Alternatively, a base plate can be connected to the column which is then installed in position on a concrete base using bolts and nuts.

design of precast columns — Precast concrete columns with protruding reinforcements

Precast columns may have corbels or nibs for supporting the beams. Alternatively, precast beam-column connections can be made using dowels or mechanical couplers.

Precast concrete columns have the following advantages over in-situ concrete construction;

Increased speed in construction since production of precast elements can commence ahead of time
Greater flexibility in project management and site planning due to off-site production capacity
Improved and higher quality of concrete, dimensions, and surface finishes
Reduction in site labour
Reduction in formwork requirement
Less wastage of materials

The design of precast reinforced concrete columns is carried out by a structural engineer and involves the following steps;

Confirm all dimensions and tolerances of the column and other members.
Analyse the structure to obtain the design bending moments, axial, and shear force
Check for column slenderness
Obtain the final design moments taking into account imperfections and second-order effects (if applicable)
Provide reinforcements to satisfy bending and axial force
Check for biaxial bending
Check for shear
Check that reinforcement provided satisfies bending and shear due to factory lifting
Check that reinforcement provided satisfies bending and shear due to site pitching
Design the connections
Detail the column as appropriate

Worked Example on the Design of Precast Columns | EN 1992-1:2004

Check the capacity of a 4.5m high 450 x 250 mm precast column to resist the action effects given below. The column is reinforced with 6 numbers of H20 mm bars. f_ck = 35 N/mm²; f_yk = 500 N/mm²; Concrete cover = 35 mm. The design has been executed using Tekla Tedds software.

Axial load and bending moments from frame analysis
Design axial load; N_Ed = 1350.0 kN
Moment about y-axis at top; M_top,y = 55.0 kNm
Moment about y-axis at bottom; M_btm,y = 22.0 kNm
Moment about z-axis at top; M_top,z = 11.4 kNm
Moment about z-axis at bottom; M_btm,z = 5.5 kNm

Column geometry
Overall depth (perpendicular to y-axis); h = 450 mm
Overall breadth (perpendicular to z-axis); b = 250; mm
Stability in the z-direction; Braced
Stability in the y-direction; Braced

Concrete details
Concrete strength class; C30/37
Partial safety factor for concrete (2.4.2.4(1)); γ_C = 1.50
Coefficient α_cc (3.1.6(1)); α_cc = 0.85
Maximum aggregate size; d_g = 20 mm

Reinforcement details
Nominal cover to links; c_nom = 35 mm
Longitudinal bar diameter; ϕ = 20 mm
Link diameter; ϕ_v = 8 mm
Total number of longitudinal bars; N = 6
No. of bars per face parallel to y-axis; N_y = 2
No. of bars per face parallel to z axis; N_z = ;3
Area of longitudinal reinforcement; A_s = N × π × ϕ² / 4 = 1885 mm²
Characteristic yield strength; f_yk = 500 N/mm²
Partial safety factor for reinft (2.4.2.4(1)); γ_S = 1.15
E_s = 200000 MPa

Column effective lengths
Effective length for buckling about y-axis; l_0y = 3500 mm
Effective length for buckling about z-axis; l_0z = 3900 mm

Effective depths of bars for bending about y-axis
Area per bar; A_bar = π × ϕ²/4 = 314 mm²
Spacing of bars in faces parallel to z-axis (centre to centre);
s_z = h – 2 × (c_nom + ϕ_v) – ϕ)/ (N_z – 1) = 172 mm
Layer 1 (in tension face); d_y1 = h – c_nom – ϕ_v – ϕ/2 = 397 mm
Layer 2; d_y2 = d_y1 – s_z = 225 mm
Layer 3; d_y3 = d_y2 – s_z = 53 mm

2nd moment of area of reinforcement about y axis;
I_sy = 2 × A_bar × [N_y × (d_y1 – h/2)²] = 3718 cm⁴
Radius of gyration of reinforcement about y-axis; i_sy = √(I_sy/A_s) = 140 mm
Effective depth about y axis (5.8.8.3(2)); d_y = h/2 + i_sy = 365 mm

Effective depths of bars for bending about z-axis
Area of per bar; A_bar = π × ϕ² / 4 = 314 mm²
Spacing of bars in faces parallel to y axis (c/c); s_y = (b – 2 × (c_nom + ϕ_v) – ϕ) / (N_y – 1) = 144 mm
Layer 1 (in tension face); d_z1 = b – c_nom – ϕ_v – ϕ/2 = 197 mm
Layer 2; d_z2 = d_z1 – s_y = 53 mm
Effective depth about z axis; d_z = d_z1 = 197 mm

Column slenderness about y-axis
Radius of gyration; i_y = h/√(12) = 13.0 cm
Slenderness ratio (5.8.3.2(1)); l_y = l_0y / i_y = 26.9

Column slenderness about z-axis
Radius of gyration; i_z = b/√(12) = 7.2 cm
Slenderness ratio (5.8.3.2(1));l_z = l_0z / i_z = 54.0

Design bending moments

Slenderness limit for buckling about y axis (cl. 5.8.3.1)
A = 0.7
Mechanical reinforcement ratio; ω = A_s × f_yd / (A_c × f_cd) = 0.429
Factor B; B = √(1 + 2ω) = 1.363
Moment ratio; r_my = M_01y / M_02y = 0.506
Factor C; C_y = 1.7 – r_my = 1.194
Relative normal force; n = N_Ed / (A_c × f_cd) = 0.706
Slenderness limit; l_limy = 20 × A × B × C_y / √(n) = 27.1

l_y < l_limy – Second order effects may be ignored

Slenderness limit for buckling about y-axis (cl. 5.8.3.1)
A = 0.7
Mechanical reinforcement ratio; w = A_s × f_yd / (A_c × f_cd) = 0.429
Factor B; B = √(1 + 2ω) = 1.363
Moment ratio; r_mz = 1.000
Factor C; C_z = 1.7 – r_mz = 0.700
Relative normal force; n = N_Ed / (A_c × f_cd) = 0.706
Slenderness limit; l_limz = 20 × A × B × C_z / √(n) = 15.9
l_z > l_limz – Second order effects must be considered

Design bending moments (cl. 6.1(4))
Design moment about y axis; M_Edy = max(M_02y, N_Ed × max(h/30, 20 mm)) = 66.8 kNm

Local second order bending moment about z-axis (cl. 5.8.8.2 & 5.8.8.3)
Relative humidity of ambient environment; RH = 50 %
Column perimeter in contact with atmosphere; u = 1400 mm
Age of concrete at loading; t₀ = 28 day
Parameter n_u; n_u = 1 + w = 1.429
n_bal = 0.4
Approx value of n at max moment of resistance; n_bal = 0.4
Axial load correction factor; K_r = min(1.0 , (n_u – n) / (n_u – n_bal)) = 0.703
Reinforcement design strain; ε_yd = f_yd/E_s = 0.00217

Basic curvature; curve_{basic_z} = ε_yd / (0.45 × d_z) = 0.0000245 mm^-1

Notional size of column; h₀ = 2 × A_c / u = 161 mm
Factor a₁ (Annex B.1(1)); a₁ = (35 MPa / f_cm)^0.7 = 0.944
Factor a₂ (Annex B.1(1)); a₂ = (35 MPa / f_cm)^0.2 = 0.984

Relative humidity factor (Annex B.1(1));
ϕ_RH = [1 + ((1 – RH/100%) / (0.1 mm^-1/3 × (h₀)^1/3)) × a₁] × a₂ = 1.838

Concrete strength factor (Annex B.1(1));
β_fcm = 16.8 × (1 MPa)^1/2 / √(f_cm) = 2.725

Concrete age factor (Annex B.1(1));
β_t0 = 1 / (0.1 + (t₀ / 1 day)^0.2) = 0.488

Notional creep coefficient (Annex B.1(1));
ϕ₀ = ϕ_RH × β_fcm × β_t0 = 2.446

Final creep development factor; (at t = ∞); β_c∞ = 1.0
Final creep coefficient (Annex B.1(1));ϕ_∞ = ϕ₀ × β_c∞ = 2.446

Ratio of SLS to ULS moments r_Mz (say) = 0.80

Effective creep ratio (5.8.4(2)); f_efz = f_∞ × r_Mz = 1.957

Factor β; β_z = 0.35 + f_ck / 200 MPa – l_z / 150 = 0.140
Creep factor; K_ϕz = max(1.0, 1 + β_z × ϕ_efz) = 1.273
Modified curvature; curve_{mod_z} = K_r × K_ϕz × curve_{basic_z} = 0.0000219 mm^-1
Curvature distribution factor; c = 10

Deflection; e_2z = curve_{mod_z} × l_0z²/c = 33.4 mm

Nominal 2^nd order moment;
M_2z = N_Ed × e_2z = 45.1 kNm

Design bending moment about z-axis (cl. 5.8.8.2 & 6.1(4))
Equivalent moment from frame analysis;
M_0ez = max(0.6 × M_02z + 0.4 × M_01z, 0.4 × M_02z) = 22.2 kNm

Design moment;
M_Edz = max(M_02z, M_0ez + M_2z, M_01z + 0.5 × M_2z, N_Ed× max(b/30, 20 mm))
M_Edz = 67.2 kNm

Moment capacity about y-axis with axial load (1350.0 kN)
Moment of resistance of concrete
By iteration:
Position of neutral axis; y = 317.8 mm

Concrete compression force (3.1.7(3));
F_yc = h × f_cd × min(l_sb × y, h) × b = 1080.6 kN

Moment of resistance;
M_Rdyc = F_yc × [h / 2 – (min(l_sb × y, h)) / 2] = 105.8 kNm

Moment of resistance of reinforcement
Strain in layer 1; ε_y1 = ε_cu3 × (1 – d_y1/y) = -0.00087
Stress in layer 1; σ_y1 = max(-1 × f_yd, E_s × ε_y1) = -174.4 N/mm²
Force in layer 1; F_y1 = N_y × A_bar × σ_y1 = -109.6 kN
Moment of resistance of layer 1; M_Rdy1 = F_y1 × (h/2 – d_y1) = 18.8 kNm

Strain in layer 2; ε_y2 = ε_cu3 × (1 – d_y2 / y) = 0.00102
Stress in layer 2; σ_y2 = min(f_yd, E_s × ε_y2) – h × f_cd = 187.4 N/mm²
Force in layer 2; F_y2 = 2 × A_bar × σ_y2 = 117.8 kN
Moment of resistance of layer 2; M_Rdy2 = F_y2 × (h/2 – d_y2) = 0.0 kNm

Strain in layer 3; ε_y3 = ε_cu3 × (1 – d_y3/y) = 0.00292
Stress in layer 3; σ_y3 = min(f_yd, E_s × ε_y3) – h × f_cd = 417.8 N/mm²
Force in layer 3; F_y3 = N_y × A_bar × σ_y3 = 262.5 kN
Moment of resistance of layer 3; M_Rdy3 = F_y3 × (h/2 – d_y3) = 45.2 kNm

Resultant concrete/steel force; F_y = 1351.2 kN
PASS – This is within half of one percent of the applied axial load

Combined moment of resistance
Moment of resistance about y axis; M_Rdy = 169.8 kNm
PASS – The moment capacity about the y axis exceeds the design bending moment

Moment capacity about z-axis with axial load (1350.0 kN)

Moment of resistance of concrete
By iteration, position of neutral axis; z = 171.9 mm
Concrete compression force (3.1.7(3)); F_zc = h × f_cd × min(l_sb × z, b) × h = 1051.9 kN
Moment of resistance; M_Rdzc = F_zc × [b / 2 – (min(l_sb × z, b)) / 2] = 59.2 kNm

Moment of resistance of reinforcement
Strain in layer 1; ε_z1 = ε_cu3 × (1 – d_z1 / z) = -0.00051
Stress in layer 1; σ_z1 = max(-1 × f_yd, E_s × ε_z1) = -102.3 N/mm²
Force in layer 1; F_z1 = N_z × A_bar × σ_z1 = -96.4 kN
Moment of resistance of layer 1; M_Rdz1 = F_z1 × (b / 2 – d_z1) = 6.9 kNm

Strain in layer 2; ε_z2 = ε_cu3 × (1 – d_z2/z) = 0.00242
Stress in layer 2; σ_z2 = min(f_yd, E_s × ε_z2) – h × f_cd = 417.8 N/mm²
Force in layer 2; F_z2 = N_z × A_bar × σ_z2 = 393.8 kN
Moment of resistance of layer 2; M_Rdz2 = F_z2 × (b/2 – d_z2) = 28.4 kNm

Resultant concrete/steel force; F_z = 1349.2 kN
PASS – This is within half of one percent of the applied axial load

Combined moment of resistance
Moment of resistance about z-axis; M_Rdz = 94.5 kNm
PASS – The moment capacity about the z-axis exceeds the design bending moment

Biaxial bending

Determine if a biaxial bending check is required (5.8.9(3))

Ratio of column slenderness ratios; ratio_l = max(l_y, l_z) / min(l_y, l_z) = 2.01
Eccentricity in direction of y axis; e_y = M_Edz/N_Ed = 49.8 mm
Eccentricity in direction of z axis; e_z = M_Edy/N_Ed = 49.5 mm

Equivalent depth; h_eq = i_y × √(12) = 450 mm
Equivalent width; b_eq = i_z × √(12) = 250 mm

Relative eccentricity in direction of y-axis; e_{rel_y} = e_y/b_eq = 0.199
Relative eccentricity in direction of z-axis; e_{rel_z} = e_z/h_eq = 0.110

Ratio of relative eccentricities;
ratio_e = min(e_{rel_y}, e_{rel_z})/max(e_{rel_y}, e_{rel_z}) = 0.552

ratio_l > 2 and ratio_e > 0.2
Therefore, biaxial bending check is required.

Biaxial bending (5.8.9(4))
Design axial resistance of section; N_Rd = (A_c × f_cd) + (A_s × f_yd) = 2732.0 kN
Ratio of applied to resistance axial loads; ratio_N = N_Ed / N_Rd = 0.494
Exponent a; a = 1.33

Biaxial bending utilisation; UF = (M_Edy/M_Rdy)^a + (M_Edz/M_Rdz)^a = 0.926 (Okay)

Shear

Design shear force; V_Ed = V_Ed,y = 25.8 kN
C_Rd,c = 0.18/γ_C = 0.12
Tension reinforcement; A_sl = N_z × π × ϕ²/4 = 942 mm²
Depth of tension reinforcement; d_v = d_z1 = 197 mm
k_shear = min(1 + (200 mm / d_v)^0.5, 2) = 2.000
Width of the cross section in tensile area; b_w = h = 450 mm
Longitudinal reinforcement ratio; r_l = min(A_sl/(b_w × d_v), 0.02) = 0.01063

Axial pressure in cross-section; σ_cp = min(N_Ed/A_c, 0.2 × f_cd) = 3.40 N/mm²
v_min = 0.035 N^0.5/mm × k_shear^3/2 × f_ck^1/2 = 0.54 N/mm²
k_1,shear = 0.15

Design shear resistance – exp. 6.2 a & b;
V_Rd,c = max(C_Rd,c × k_shear × (100 N²/mm⁴ × r_l × f_ck)^1/3, v_min) × b_w × d_v + k_1,shear × σ_cp × b_w × d_v = 112.7 kN
V_Ed / V_Rd,c = 0.23
PASS – Design shear resistance exceeds design shear force

Factory Lifting Check

Precast element details
Total length of column; L_element = 4500 mm
Distance between lifting points; L_lift = 2500 mm
Lifting load coefficient; f_lifting = 1.50
Permanent load factor; γ_G = 1.35
Formwork adhesion force; q_formwork = 2.0 kN/m²
Self weight of precast element; w_{self_precast} = b × h × ρ_conc × g_acc + q_formwork × b = 3.3 kN/m

Lifting check (positive moment)
Design bending moment; M = γ_G × f_lifting × (w_{self_precast} × L_lift² /8 – w_{self_precast} × ((L_element – L_lift) / 2)²/2) = 1.9 kNm

Effective depth of tension reinforcement; d = 397 mm
Redistribution ratio; d = 1.000
K = M / (b × d² × f_ck) = 0.002
No compression reinforcement is required

Lever arm; z = min(0.5 × d × (1 + (1 – 2 × K / (h × a_cc / γ_C))^0.5), 0.95 × d) = 377 mm
Depth of neutral axis; x = 2 × (d – z)/l_sb = 50 mm

Area of tension reinforcement required; A_s,pos = M / (f_yd × z) = 11 mm²
Tension reinforcement provided; 2H20 mm (A_s,prov = 628 mm²)
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm/f_yk, 0.0013) × b × d = 149 mm²
Maximum area of reinforcement – cl.9.2.1(3); A_s,max = 0.04 × b × h = 4500 mm²
Required area of reinforcement; A_s,req = 149 mm²
A_s,req / A_s,prov = 0.24 (okay)

Lifting check (negative moment)
Design bending moment; M = γ_G × f_lifting × w_{self_precast} × ((L_element – L_lift) / 2)² / 2 = 3.3 kNm
Effective depth of tension reinforcement; d = 397 mm
Redistribution ratio; d = 1.000
K = M / (b × d² × f_ck) = 0.003

Area of tension reinforcement required; A_s,neg = M / (f_yd × z) = 20 mm²
Tension reinforcement provided; 2H20 mm (A_s,prov = 628 mm²)
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm/f_yk, 0.0013) × b × d = 149 mm²
Maximum area of reinforcement – cl.9.2.1(3); A_s,max = 0.04 × b × h = 4500 mm²
Required area of reinforcement; A_s,req = 149 mm²
A_s,req / A_s,prov = 0.24 (Okay)

Lifting check (Shear)
Design shear force at critical shear plane;
V_Ed = γ_G × f_lifting × w_{self_precast} × max(L_lift / 2, (L_element – L_lift) / 2) = 8.2 kN

C_Rd,c = 0.18/γ_C = 0.12

Tension reinforcement; A_sl = N_y × π × ϕ² / 4 = 628 mm²
Depth of tension reinforcement; d_v = d_y1 = 397 mm
k_shear = min(1 + (200 mm / d_v)^0.5, 2) = 1.710

Width of the cross section in tensile area; b_w = b = 250 mm

Longitudinal reinforcement ratio; ρ_l = min(A_sl / (b_w × d_v), 0.02) = 0.00633
v_min = 0.035 N^0.5/mm × k_shear^3/2 × f_ck^1/2 = 0.43 N/mm²

Design shear resistance – exp. 6.2 a & b;
V_Rd,c = max(C_Rd,c × k_shear × (100 N²/mm⁴ × ρ_l × f_ck)^1/3, v_min) × b_w × d_v
V_Rd,c = 54.3 kN
V_Ed / V_Rd,c = 0.15 (This is okay)

On-site Pitching Check

Precast element details
Total length of column; L_element = 4500 mm
Distance to the pitching point; L_pitch = 1800 mm
Distance from pitching point to end of column;L_end = 2700 mm
Lifting load coefficient; f_pitching = 1.25
Permanent load factor; g_G = 1.35
Self weight of precast element; w_{self_precast} = b × h × ρ_conc × g_acc = 2.8 kN/m

Lifting check (positive moment)
Design bending moment (at 3750 mm);
M = g_G × f_pitching × w_{self_precast} × L_element² / (2 × L_end) × (0.25 × L_element²/L_end – L_pitch) = 1.3 kNm

Effective depth of tension reinforcement; d = 397 mm
Redistribution ratio; d = 1.000
K = M / (b × d² × f_ck) = 0.001

Area of tension reinforcement required; A_s,pos = M / (f_yd × z) = 8 mm²
Tension reinforcement provided; 2H20 mm (A_s,prov = 628 mm²)
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm/f_yk, 0.0013) × b × d = 149 mm²
Maximum area of reinforcement – cl.9.2.1(3); A_s,max = 0.04 × b × h = 4500 mm²
Required area of reinforcement; A_s,req = 149 mm²
A_s,req / A_s,prov = 0.24 (Okay)

Lifting check (negative moment)
Design bending moment; M = g_G × f_pitching × w_{self_precast} × L_pitch ² / 2 = 7.5 kNm
Effective depth of tension reinforcement; d = 397 mm
Redistribution ratio; d = 1.000

K = M / (b × d² × f_ck) = 0.006

Area of tension reinforcement required; A_s,neg = M / (f_yd × z) = 46 mm²

Tension reinforcement provided; 2H20 mm (A_s,prov = 628 mm²)
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm/f_yk, 0.0013) × b × d = 149 mm²
Maximum area of reinforcement – cl.9.2.1(3); A_s,max = 0.04 × b × h = 4500 mm²
Required area of reinforcement; A_s,req = 149 mm²
A_s,req / A_s,prov = 0.24 (Okay)

Lifting check (Shear)
Design shear force at critical shear plane;
V_Ed = g_G × f_pitching × w_{self_precast} × max(L_pitch, abs(L_element – 0.5 × L_element² / L_end), abs(L_end – (L_element – 0.5 × L_element² / L_end))) = 9.1 kN

C_Rd,c = 0.18/γ_C = 0.12
Tension reinforcement; A_sl = N_y × π × ϕ² / 4 = 628 mm²
Depth of tension reinforcement; d_v = d_y1 = 397 mm
k_shear = min(1 + (200 mm / d_v)^0.5, 2) = 1.710

Width of the cross section in tensile area; b_w = b = 250 mm

Longitudinal reinforcement ratio; ρ_l = min(A_sl / (b_w × d_v), 0.02) = 0.00633
v_min = 0.035 N^0.5/mm × k_shear^3/2 × f_ck^1/2 = 0.43 N/mm²

Design shear resistance – exp. 6.2 a & b;
V_Rd,c = max(C_Rd,c × k_shear × (100 N²/mm⁴ × ρ_l × f_ck)^1/3, v_min) × b_w × d_v
V_Rd,c = 54.3 kN
V_Ed / V_Rd,c = 0.17 (Okay)

Connection

The connection of the column can be designed and checked depending on the method adopted.

Single Wheel Load Distribution on Bridge Decks

By

Ubani Obinna

-

April 27, 2021

Moving traffic is the major live load (variable action) on bridge decks. In the design of bridges, it is very important to consider the global and local effects of moving traffic loads on the bridge. While global effects can be used to distribute traffic wheel load to the girders (in the case of beam and slab bridges), local verification is very important in the design of deck slabs especially for bending moment and shear. In this article, we are going to show how to distribute single wheel load to bridge deck slabs.

In BS 5400, single HA load (with a value of 100 kN) is normally used for all types of local effects verification. This load is normally assumed to act on a square contact area of 300 mm x 300 mm to give a pressure of 1.11 N/mm² on the surface where it is acting. For 45 units of HB load, a single wheel load of 112.5 kN can be considered where necessary.

In EN 1991-2 (Eurocode specification for bridges), the local wheel load is represented by Load Model 2 (LM2) which consists of a single axle load with a magnitude of 400 kN (inclusive of the dynamic amplification factor). The single axle consists of two wheels (200 kN each) spaced at a distance of 2m (centre to centre). This load model is intended to be used for local verification only and should be considered alone on the longitudinal axis of thebridge. Unless otherwise specificied, each wheel is assummed to act on a rectangular area of 600 x 350 mm.

To apply single wheel load on bridge decks for local verification, the wheel load is placed at the most adverse location on the bridge deck, and the contact pressure distributed to the neutral axis of the bridge deck. For all practical purposes, the neutral axis is normally taken at the mid-section of the bridge deck. For distribution of wheel load through concrete deck slabs, a ratio of 1:1 is normally adopted (dispersal angle of 45 degrees), while for distribution through asphalt surfacing, a dispersal ratio of 1:2 is normally adopted.

When the contact pressure has been obtained (together) with the dimensions of the contact area, the effects of the wheel load on the slab can be assessed using any suitable method. For manual analysis, Pigeaud’s curve can be used to obtain the design bending moments. For finite element analysis, the contact pressure can be applied as a patch load (partially distributed load) on a slab surface.

To show how this is done, let us consider a worked example.

Worked Example

Obtain the design moments in an interior panel of deck slab of the bridge system shown below due to the effect of a single HA wheel load on the bridge deck. The deck slab is overlain with a 75mm thick asphalt surfacing.

The dimensions of an interior panel of the bridge deck are as follows;

Length (l_y) = 4.5 m
Width (l_x) = 1.8
Support condition: Continuous over all supports

Design wheel load = 100 kN
Contact area = 300 x 300 mm

The contact pressure = 100 kN/(0.3 x 0.3)m = 1111.11 kN/m²

A 2(V):1(H) load distribution through the 75 mm thick asphalt to the surface of the deck slab will give a pressure of 711.11 kN/m²acting on a square contact area of 375 x 375 mm.

A further 1:1 load distribution from the surface of the concrete deck slab to the neutral axis will give a pressure of 256 kN/m² acting on a square contact area of 625 x 625 mm.

Therefore;
a_x = a_y = 625 mm

To enable us use the Pigeaud’s curves;
k = l_y/l_x = 4.5/1.8 = 2.5
a_x/l_x = 0.625/1.8 = 0.347
a_y/l_y = 0.625/4.5 =0.138

From Table 55 of Reynolds and Steedman (10th Edition),
α_x4 = 0.169
α_y4 = 0.0964

Poisson’s ration v = 0.2

Transverse bending moment M_x = F(α_x4 + vα_y4) = 100 x (0.169 + 0.2 x 0.0964) = 18.828 kNm/m
Longitudinal bending moment M_y = F(vα_x4 + α_y4) = 100 x (0.2 x 0.169 + 0.0964) = 13.02 kNm/m

Taking account of suggested allowances for continuity in the transverse direction;
M_x1 = -0.25M_x = 0.25 x 18.828 = -4.707 kNm

These are therefore the bending moments due to the HA wheel load on the deck slab

Design of Inclined Columns | Slanted Columns

By

Ubani Obinna

-

April 22, 2021

Inclined or slanted columns are columns that are leaning at an angle away from perfect verticality (90 degrees to the horizontal). This is usually intentional and not due to imperfection from materials or construction. The degree of inclination can vary depending on the designer’s intentions. In this article, we are going to review the analysis and design of inclined columns using standard design codes.

Inclined columns can be introduced into a building to serve architectural or structural functions. The design of an inclined column is like the design of any other column but with special attention paid to the changes in stresses due to the eccentricity of the axial load on the column. In some cases, inclined columns can be more susceptible to second-order effects than perfectly vertical columns.

When a column in a structure is perfectly straight, first-order bending moment and other internal stresses are induced due to the externally applied loads. However, when a column is inclined at an angle to the beams or floor that it is supporting, changes in bending moments are observed due to the eccentricity of the axial load with respect to the longitudinal axis column. Such effects can be investigated for pinned supports or fixed supports.

In the design of inclined columns (say in reinforced concrete or steel), it is usually very sufficient to analyse the structure and obtain the design internal forces (bending moment, shear, and axial force) using first-order linear analysis. However, when required, the analysis can be extended to second-order non-linear analysis to account for secondary effects which may affect the stability of the column.

To verify the effect of inclination on the design of columns, let us investigate the following cases of a column with a fixed base in reinforced-concrete construction.

Dimensions of beam = 600 x 300 mm
Dimensiond of column = 300 x 300 mm
Ultimate load (factored load) on the beam = 70 kN/m

When the frames were analysed using first-order linear analysis, the following results were obtained;

CASE A

Maximum column design axial force = 220.219 kN
Column design shear force = 22.880 kN
Column design moment = 61.313 kNm

CASE B

Maximum column design axial force = 241.712 kN
Column design shear force = 19.608 kN
Column design moment = 56.624 kNm

CASE C

Maximum column design axial force = 226.991 kN
Column design shear force = 20.892 kN
Column design moment = 59.247 kNm

From the analysis results, it could be seen that case A gave the highest column moment but the least axial force, while case B gave the lowest column design moment but highest column axial force. The design of the column can then be carried out for each of the cases presented.

Strengthening of Flat Slabs with Cut-Out Openings

By

Ubani Obinna

-

April 18, 2021

A flat slab is a type of reinforced concrete floor system where the slab is supported directly by columns, instead of beams. This system offers many advantages such as reduced headroom, flexibility in the distribution of services, reduced formwork complexity during construction, etc. Strengthening of flat slabs may become necessary especially when openings are introduced.

According to recent research (2021) from the Department of Civil Engineering, University of Wasit, Iraq, openings can be created in a slab after construction in order to pass the new services that satisfy the occupants’ desires, such as water or gas piping, ventilation, electricity, elevator, and staircase. The creation of openings in a flat slab has been found to significantly reduce the structural integrity of the floor.

Flat slabs can undergo brittle failure due to punching shear around the columns when the thickness of the slab is insufficient and/or when punching shear reinforcement is not provided. According to Hussein et al (2021), the possibility of punching shear failure increases by creating openings next to the columns due to removing a substantial amount of concrete from the critical section around columns, which is responsible for resisting the punching force. This is usually accompanied by the cutting of several flexural reinforcement bars.

This undesirable impact of openings gets more severe when they are not taken into account during the design, or they are installed after the construction of slabs. The opening influence could be minimized when they are positioned away from the columns, or the slab depth is increased.

From many research works carried out on flat slabs with openings, the following conclusions were reached. The full references and details of the research works can be found in Huessein et al., (2021).

For openings of the same size, the strength reduction was more significant as the number of openings around the column increased.
The influence of openings on the strength of slabs is greater when they are constructed in large sizes, especially when the size of the opening is greater than the size of the corresponding column.
The impact of the opening could be reduced by shifting them far away from the supporting columns. Two far openings were found to be less detrimental than ones adjacent to column’s faces.
Openings located at column corners reduced the strength of slabs more than openings placed in front of the column
For openings of the same number and size, the arrangement of openings as the letter (L) around the column caused a lesser drop in the slabs’ punching strength.

In general, the reported maximum reduction in the slabs’ strength due to openings was about 29% relative to the solid equivalents.

It, therefore, makes sense to strengthen flat slabs with cut-out opening, especially when the effect of the opening was not taken into account during the design. Therefore, Hussein et al (2021) decided to investigate the effects of strengthening flat slabs with cut-out openings using different materials. The strengthening techniques investigated in the study were Carbon Fiber Reinforced Polymer (CFRP), steel plates, steel bars, and near-surface mounted Engineered Cementations Composite (ECC) with steel mesh. The findings were published in Case Studies of Construction Materials (Elsevier).

brrt — Fig 1: Details of the specimens used in the investigation (Hussein et al, 2021)

In order to investigate how strengthening can retrieve the lost mechanical strength of slab with openings, six reinforced concrete slabs were prepared with dimensions of 1300 x 1300 x 120 mm. One of them was reference without opening, whereas the others contained a square edge opening of 350 mm side. For slabs with openings, one specimen was the control left without strengthening, and the remaining four were strengthened utilizing various methods stated above.

drrt — *Fig 2: Details of the investigation techniques used in the research (Hussein et al, 2021)*

The details of the preparation of the specimens can be found in Hussein et al, (2021). After the curing process completed, the specimens were tested under the influence of uniform loading. The samples were moved into a testing steel rig and rested at their corners on supporting steel plates, measuring 200 x 200 x 30 mm to simulate the real supporting columns. Nevertheless, the rotations around these plates were not restricted. The uniform loading was applied incrementally with the help of a hydraulic jack up to failure.

The major results of the experiment are shown below (Hussein et al., 2021);

Based on the experiments by Hussein et al (2021) conducted on six reinforced concrete slabs, including edge openings and strengthened using four various methods, tested under uniformly distributed loads, the following points can be put forward as the essential outcomes;

1. Installation of an opening at the edge of a flat slab resting on four columns and subjected to uniformly distributed load did not change the failure modes. However, the losses in strength, ductility, and toughness were 20.6 %, 16.2 %, and 38 %, respectively when compared with the solid slab.
2. Among the four adopted methods of strengthening, the use of embedded steel bars and EEC altered the failure mode from brittle due to punching shear to more ductile combined flexural-shear modes.
3. The effectiveness of retrofitting techniques in restoring the lost mechanical properties of flat slabs due to openings was found to relate directly to the bond strength between the strengthening materials and the slab surface. Accordingly, the embedded steel bars technique was the most efficient, whereas the ECC was the least efficient method.
4. The strengthening materials should be extended as long as possible from the opening edge in order to achieve a sufficient development length. Hence, the retrofitting materials’ ultimate strength can arrive, and the debonding can be delayed.
5. None of the strengthening methods managed to restore the total missing strength of slabs due to openings. The use of embedded steel bars achieved the most significant upgrading in the slab strength, about 22.2 % over that of the slab with opening and 3 % below the solid slab’s strength.

References
Hussein M. J., Jabir H. A., Al-Gasham T. S. (2021): Retrofitting of reinforced concrete flat slabs with cut-out edge opening. Case Studies in Construction Materials 14 (2021) e00537 https://doi.org/10.1016/j.cscm.2021.e00537

Disclaimer
The contents of the above-cited research work have been presented on www.structville.com because it is an open-access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/)

How to Distribute HB Live Load to Bridge Girders

By

Ubani Obinna

-

April 17, 2021

In the UK, HB loads are used to represent abnormal traffic on a bridge deck. Even though the use of HA and HB loads as traffic actions has been replaced with Load Models 1 to 4 of EN 1991-2, the use of HB load is still applied in the design of bridges especially in Nigeria. It is very important, therefore, to know how to analyse bridge girders (in beam and slab bridges) for HB loads in order to obtain the design forces due to moving traffic.

There are many methods through which the effects of traffic wheel load can be obtained on the longitudinal components of a bridge deck. Some of these methods are;

Finite element analysis
Grillage analysis, and
Simplified methods

The simplified methods of analysing bridge decks are known as the manual methods. While bridge design all over the world is majorly carried out using computers, the simplified methods (which are pretty quick to apply) can be used for quick checks or for preliminary and/or detailed design. The difference between the results from computer methods and simplified methods is often not very wide.

The basis of many simplified/manual methods is the distribution coefficient methods, e.g. Morice and Little (1956). These methods can be applied manually to obtain the values of various load effects at any reference point on a transverse section of the bridge.

For most of the load distribution coefficient methods, a right simply supported bridge is idealized as an orthotropic plate whose load distribution characteristics are governed by two dimensionless parameters α and θ. These parameters depend largely on the geometry of the bridge deck (which is usually known prior to the analysis).

In North America, the simplified method of bridge deck analysis is widely used. Unlike the distribution coefficient methods, the simplified methods of bridge analysis provide only the maximum action effect for the purposes of design. As a result, a lesser computational effort is required when using this method compared with the distribution coefficient method. These simplified methods are permitted by the current and past design codes, being the AASHTO Specifications (1998, 2010), the CSA Code (1988), the Ontario Highway Bridge Design Code (1992), and the Canadian Highway Bridge Design Code (2000, 2006).

The results from the North American simplified methods are reliable and can be obtained rather quickly. It is important to note that the analysis depends on the specification of the magnitude and placement of the design live loads, and accordingly are not always transportable between the various design codes.

The D-Method of Bridge Deck Analysis

In the simplified method of analysis, a longitudinal girder, or a strip of unit width in the case of slabs, is isolated from the rest of the structure and treated as a one-dimensional beam. This isolated beam is subjected to loads comprising one line of wheels of the design vehicle multiplied by a load fraction (S/D), where S is the girder spacing and D, having the units of length, has an assigned value for a given bridge type (Bakht and Mufti, 2015).

concept of the D method bridge deck — **Fig 1**: Transverse distribution of longitudinal moment intensity (Bakht and Mufti, 2015)

The concept of the factor D can be explained with reference to the figure above, which shows schematically the transverse distribution of live load longitudinal moment intensity in a slab-on-girder bridge at a cross-section due to one vehicle with two lines of wheels.

The intensity of longitudinal moment, having the units of kN.m/m, is obtained by idealizing the bridge as an orthotropic plate. A little consideration will show that the maximum girder moment, M_g, for the case under consideration occurs in the second girder from the left.

The moment in this girder is equal to the area of the shaded portion under the moment intensity curve. If the intensity of maximum moment is M_x(max) then this shaded area is approximately equal to SM_x(max), so that:

M_g ≃ SM_x(max) —— (1)

It is assumed that the unknown quantity M_x(max) is given by:

M_x(max) = M/D —— (2)

where M is equal to the total moment due to half a vehicle, i.e., due to one line of wheels. Substituting the value of M_x(max) from Eq. (2) into Eq. (1):

M_g = M(S/D) —— (3)

Thus if the value of D is known, the whole process of obtaining longitudinal moments in a girder is reduced to the analysis of a 1-dimensional beam in which the loads of one line of wheels are multiplied by the load fraction (S/D).

Simplified Method of Analysis for HB Load

A simplified method for two-lane slab-on-girder bridges in Hong Kong was presented by Chan et al. (1995). This was based on HB loads that are popularly used for bridge design in the UK.

The HB loading comprises four axles, with four wheels in each axle. The weights of the wheels are governed by the units of the HB loading. In Great Britain,we have 30 units, 37.5, and 45 units of HB. Each unit of HB is equal to 10 kN. For instance, for 45 units of HB load, each axle has a total load of 450 kN, while each wheel has a load of 112.5 kN. It should, however, be noted that the units of the design loading does not affect the simplified method presented by Chan et al. (1995).

HB wheel load configuration — **Fig 2**: HB wheel load arrangement

According to Chan et al (1995), the longitudinal flexural rigidity per unit width D_x, of slab-on-girder bridges in Hong Kong lies between two bounds defined by the following equations.

D_x = 48000L + 5100L² (upper bound) —— (4)
D_x = 2000L + 3650L² (lower bound) —— (5)

where the span of the bridge L is in metres and D_x in kN.m.

From a study of a large number of slab-on-girder highway bridges in Hong Kong, it was determined that the ranges of the various parameters which influence the transverse load distribution characteristics of a bridge are as follows.

(a) The deck slab thickness, t, varies between 150 and 230 mm, with the usual value being 200 mm.
(b) The centre-to-centre spacing of girders, S, varies between 0.2 and 2.0 m, with the usual value being 1.0 m.
(c) The lane width, W_e, varies between 3.2 and 3.8 m, with the usual value being 3.5 m.
(d) The vehicle edge distance, VED, being the transverse distance between the centre of the outermost line of wheels of the HB loading and the nearest longitudinal free edge of the bridge, varies between 0.75 and 5.00 m, with the usual value being 1.00 m.

From the above observations, the following values of the various parameters were adopted for the developmental analyses conducted for developing the simplified method: t = 200 mm; S = 1.0 m; W_e = 3.5 m; and VED = 1.00 m. In addition, it was assumed that the deck slab overhang beyond the outer girders was 0.55 m. Bridges with spans of 10, 20, 30 and 40 m were selected for the developmental analyses.

Chan et al. (1995), having plotted the values of D from the above analyses against the span length L, found that these values of D are related to L according to the following equations with a reasonable degree of accuracy, provided that the design value of D, i.e. D_d, is corrected.

For internal girders having L < 25 m:
D = 1.2 – 3.5/L —— (6)

For internal girders having L ≥ 25 m:
D = 1.06 —— (7)

For external girders having L < 30 m:
D = 0.95 + 2.1/L —— (8)

For external girders having L ≥ 30 m:
D = 1.03 —— (9)

The correcting equation for obtaining D_d is as follows;

D_d = D(1 + µC_w/100) —— (10)

Where;
µ = (3.5 – W_e)/0.25 —— (11)

The values of cw for internal and external girders can be read from the chart below;

The following conditions must be met for applying the simplified method.

The value of D_x lies between the upper and lower bound values given by Eqs. (4) and (5), respectively.
The bridge has two design lanes
The width is constant or nearly constant, and there are at least three girders in the bridge.
The skew parameter ε = (S tan ψ)/L does not exceed 1/18 where S is girder spacing, L is span and ψ is the angle of skew.
For bridges curved in plan, L²/bR does not exceed 1.0, where R is the radius of curvature; L is span length; 2b is the width of the bridge.
The total flexural rigidity of transverse cross-section remains substantially the same over at least the central 50% length of each span.
Girders are of equal flexural rigidity and equally spaced, or with variations from the mean of not more than 10% in each case.
The deck slab overhang does not exceed 0.6S, and is not more than 1.8 m.

The steps for applying the simplified method in the analysis of bending moment due to HB load on a bridge girder are;

Calculate the value of D from the relevant of Eqs. (6, 7, 8, and 9); for simply supported spans, L is the actual span length, and for continuous span bridges, the effective L for different spans should be obtained.
For the design lane width, W_e, obtain μ from Eq. (11) and C_w from the chart and thereafter obtain D_d from Eq. (10).
Isolate one girder and the associated portion of the deck slab, and analyse it by treating it as a one-dimensional beam under one line of wheels of the HB loading. The moment thus obtained at any transverse section of the beam is designated as M.
For any of the internal and internal girders, obtain the maximum moment at the transverse section under consideration by multiplying M with (S/D_d), where D_d is as obtained in Step (b) for the relevant internal or external girders.

Worked Example

Obtain the maximum sagging moment in an interior girder of a one-span bridge deck carrying 45 units of HB load. The configuration of the bridge deck is shown below;

Actual span lengths, L = 18.0 m
Bridge width, 2b = 8.7 m
Carriageway width = 7.2 m
Number of notional lanes = 2
Design lane width (notional lane width) = W_e = 3.6 m
Deck slab thickness = 0.20 m
Girder spacing S = 1.80 m
Deck slab overhang = 0.75 m
Vehicle edge distance VED = 1.35 m

HB line beam analysis

Isolating a girder of the bridge and placing a line of HB wheel load on it to determine the maximum moment. The wheel load position that will produce the maximum moment is shown below;

HB load line beam analysis — **Fig 5**: Line beam analysis of an isolated girder under one line of HB load

When analysed;

M= 1203.75 kNm

From equations (6) and (8);

Internal girder (L < 25 m)
D = 1.2 – 3.5/L = 1.2 – 3.5/18 = 1.0055

External girder (L < 25 m)
D = 0.95 + 2.1/L = 0.95 + 2.1/18 = 1.067

Applying a correction factor to the value of D;
D_d = D(1 + µC_w/100)

Where;
µ = (3.5 – W_e)/0.25 = (3.5 – 3.6)/0.25 = -0.4

For internal girder;
C_w = 6L/40 = (6 x 18)/40 = 2.7%
D_d = 1.0055[1 – (0.4 x 2.7)/100] = 0.9946
Load Fraction S/D = 1.8/0.9946 = 1.8097

For external girder;
C_w = 14 %
D_d = 1.067[1 – (0.4 x 15)/100] = 1.00298
Load Fraction S/D = 1.8/1.00298 = 1.794

Therefore the maximum moment in the internal girder = 1.8097 x 1203.75 = 2178 kNm
Therefore the maximum moment in the external girder = 1.794 x 1203.75 = 2159 kNm

Kindly check for the accuracy of this answer using any method available to you and post your findings in the comment section.

Disclaimer:
Most of the procedures presented here are as described by Bakht and Mufti (2015) with minor alterations after studying the original article cited (Chan et al, 2015). The copyright to those contents, therefore, belongs to the original owners. The worked example used to illustrate the procedure was however developed by Structville.com

References
Bakht B., Mufti A. (2015): Bridge Analysis, Design, Structural Health Monitoring, and Rehabilitation. Springer
Chan THT, Bakht B, Wong MY (1995): An introduction to simplified methods of bridge analysis for Hong Kong. HKIE Trans 2(1):1–8
Morice P. B. and Little G. (1956): The Analysis of Right BridgeDecks Subjected to Abnormal Loading. Cement and Concrete Association, London, Report Db 11

Guidelines for Writing Volume 1 – ‘Post-Graduate Experience’

Post Graduate Training – Titled Experience

Project Title

Organisation Name

Statement of Project Objective

Project Duration and Position Held

Description of Experience

Pictures

Guidelines for Writing Volume 2 – ‘NSE Technical Report on Two Selected Projects’

Introduction

Justification of the Project

Preliminary Studies/Investigations

Design Considerations/Criteria

Standards and Specifications

Methodology/Design Calculations

Drawings

Preparation of BEME

Types of Earth Dams

Homogeneous Embankment Type

Zone Embankment Type

Diaphragm Type Embankment

Methods of Construction

Hydraulic-fill Method

Rolled-fill Method

Failure of Earth Dams

Hydraulic Failures

Seepage Failures

Structural Failures

Detailing Information

The Minimum Area of Reinforcement for Solid Slabs

Bar spacing

Anchorage and Lapping of Bars

Simplified Curtailment Rules for Reinforcement

Notation for Locating Layers of Reinforcement

Reinforcement Bars and Indicator Lines

Edge reinforcement

The Trapezoidal Rule for Numerical Integration

Composite Trapezoidal Rule

The Simpson’s Rule for Numerical Integration

Composite Simpson’s Rule

Worked Example

Worked Example on the Design of Precast Columns | EN 1992-1:2004

Design bending moments

Biaxial bending

Shear

Factory Lifting Check

On-site Pitching Check

Connection

Worked Example

The D-Method of Bridge Deck Analysis

Simplified Method of Analysis for HB Load

Worked Example

EDITOR PICKS

POPULAR POSTS

POPULAR CATEGORY

ABOUT US

FOLLOW US